Import Question JSON

Current Question (ID: 12536)

Question:
In a leaf, oval cells with large intercellular spaces and radially arranged columnar cells without intercellular spaces are placed respectively towards
Options:
  • 1. Adaxial and abaxial epidermis
  • 2. Abaxial and adaxial epidermis
  • 3. Abaxial and abaxial epidermis
  • 4. Lower and abaxial epidermis
Solution:
In a leaf, oval cells with large intercellular spaces are typically found in the abaxial epidermis, which is the lower surface of the leaf. These cells are known as stomata, which are specialized structures responsible for gas exchange (such as oxygen and carbon dioxide) and water vapor regulation in plants. On the other hand, radially arranged columnar cells without intercellular spaces are typically found in the adaxial epidermis, which is the upper surface of the leaf. These cells are known as palisade mesophyll cells and are responsible for photosynthesis, as they contain chloroplasts and are the main site of photosynthetic activity in the leaf. Therefore, option 2 ("Abaxial and adaxial epidermis") is the correct answer, as it correctly identifies the placement of the described cell types in the leaf. Option 1 ("Adaxial and abaxial epidermis") and option 4 ("Lower and abaxial epidermis") are not correct, as they have the placement of the cell types reversed. Option 3 ("Abaxial and abaxial epidermis") is not correct, as it suggests that both types of cells are found on the lower surface of the leaf, which is not accurate.

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}